We only need to prove the “only if” part of the statement.
Assume that \(S \setminus \bigcup _{i=1}^n \pi ^{\prime }_i (Y^{\prime }_i(K))\)  is dense for all finite surjective separable \(\pi ^{\prime }_i \colon Y^{\prime }_i \rightarrow X\)  as in the statement of the lemma. We then have to show that \(S\)  is not thin, i.e., given integral \(K\) -varieties \(Y_1,\ldots ,Y_n\)  and \(\pi _i \colon Y_i \rightarrow X\)  dominant generically finite separable morphisms of degree \(\deg \pi _i \ge 2\) , we have to show that there exists a dense set of points \(x \in S\)  of \(X\)  such that \(Y_{i,x}(K) = \emptyset \)  for all \(i\) .
If any of the \(Y_i\)  is not geometrically irreducible, then \(Y_i(K)\)  is not dense in \(Y_i\) , and therefore \(\pi _i(Y_i(K))\)  is not dense in \(X\) , so we may and do assume all \(Y_i\)  to be geometrically irreducible.
Since the morphisms \(\pi _i\)  are separated and of finite type, we can use Nagata's compactification theorem (see [1]}, [2]}, or [3]} for a more modern proof) to obtain an open immersion \(Y_i \hookrightarrow Y_i^{\prime }\)  over \(X\) , where the morphism \(\pi _i^{\prime } \colon Y_i^{\prime } \rightarrow X\)  is proper.
If \(Y^{\prime }_{i,x}(K) = \emptyset \)  for some \(x\in S\) , then this implies \(Y_{i,x}(K) = \emptyset \) , so it suffices to show that \(S \setminus \bigcup _{i=1}^n \pi ^{\prime }_i (Y^{\prime }_i(K))\)  is dense in \(X\) .
Let
\(Y^{\prime }_i\overset{f_i}{\longrightarrow } X_i\overset{g_i}{\longrightarrow } X\) 
be the Stein factorization of \(\pi ^{\prime }_i\) , where \(f_i\)  is proper and has geometrically connected fibers, and \(g_i\)  is finite. Since \(\pi ^{\prime }_i\)  is dominant and proper (which implies closed), it is also surjective, so \(g_i\)  must be surjective too.
If \(\eta _i\)  denotes the generic point of \(X_i\) , then \(Y^{\prime }_{i,\eta _i}\)  is finite and connected, i.e. an isolated point. Since \(X_i\)  is normal, \(f_i\)  is birational by Zariski's Main Theorem.
Thus, we get
\(\deg (X_i \rightarrow X) = \deg (Y^{\prime }_i \rightarrow X) = \deg (Y_i \rightarrow X) \ge 2.\) 
The \(X_i \rightarrow X\)  are finite surjective separable of degree at least two, so that by our assumption there is a dense set of \(K\) -points \(x\in S\)  of \(X\)  such that none of the \(X_{i,x}\)  have a \(K\) -point, hence neither do the \(Y^{\prime }_{i,x}\) , which is what we had to show.
